Let $f$ be a transformation from $\mathbb{R}^3$ to $\mathbb{R}^3$. Its Jacobian matrix is given below. $J(f) = \begin{bmatrix} 2xy & x^2 & 0 \\ \\ 0 & -1 & 1 \\ \\ 1 & 0 & 1 \end{bmatrix}$ Find the Jacobian determinant of $f$. $|J(f)| = $ How will $f$ expand or contract space around the point $\left( 1, \dfrac{1}{4}, 2 \right)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Leave it the same (Choice B) B Expand it finitely (Choice C) C Contract it finitely (Choice D) D Contract it infinitely
The Jacobian determinant is the determinant of the Jacobian matrix. It represents the factor by which the transformation $f$ expands or contracts volume around a certain input. $\begin{aligned} |J(f)| &= \det \left( \begin{bmatrix} 2xy & x^2 & 0 \\ \\ 0 & -1 & 1 \\ \\ 1 & 0 & 1 \end{bmatrix} \right) \\ \\ &= 2xy(-1 - 0) - x^2(0 - 1) + 0(0 + 1) \\ \\ &= x^2 - 2xy \end{aligned}$ If we evaluate $|J(f)|$ at $\left( 1, \dfrac{1}{4}, 2 \right)$, we get $\dfrac{1}{2}$. Because the Jacobian determinant here has an absolute value less than $1$ but not $0$, we can conclude that $f$ will finitely contract the space around $\left( 1, \dfrac{1}{4}, 2 \right)$. To recap, the Jacobian determinant of $f$ is $x^2 - 2xy$, and $f$ will finitely contract the space around the point $\left( 1, \dfrac{1}{4}, 2 \right)$.